Since we don't know the population standard deviation (only the sample SD) we should use the t-distribution here. Our degrees of freedom will be the sample size, n=100.
First, find the standard error of the sample mean. This will be:
SEM = sample sd / sqrt(n), where n is the sample size.
SEM = 0.4 / sqrt(100) = 0.04.
Next, find the t value associated with our measurement of 16.1. This will be:
(measurement - pop mean) / SEM
t = (16.1-16)/0.04
Now we can go in two different ways (they'll give you the same answer) to test the hypothesis. One, find the probability of seeing a lower t-value than 2.5 for our degrees of freedom:
P(t<2.5, df = 100) = 0.993
Since this is greater than (1 - alpha/2), we reject the null hypothesis. We'd do the same if it was less than alpha/2, i.e. the null hypothesis would be accepted for alpha/2 < P(t) < 1-alpha/2.
Alternatively, you can find the critical t for a 95% confidence interval (100% - alpha) using the inverse. Remember it's two-tailed, so the probability you use is (1-alpha/2), not (1-alpha) since you need the 95% range in the middle so it stretches from 2.5% to 97.5%:
(t | P=0.975, df = 100) = t.inv(0.975,100) = 1.984
Since the magnitude of the t-stat we found (2.5) is greater than the critical t for the test (1.984), we reject the null hypothesis (note that if we found -2.5, we'd also reject since the magnitude of -2.5 is 2.5).
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